Question: $\int (3 x^5 - x^3 +6)\,dx=$ $+C$
Explanation: We can use the sum rule and the constant multiple rule for indefinite integrals: $\begin{aligned} &\int [f(x)+g(x)]dx=\int f(x)\,dx+\int g(x)\,dx \\\\\\ &\int k\cdot f(x)= k\cdot\int f(x)\,dx \end{aligned}$ Using the sum and the constant multiple rules, we can rewrite our integral as follows: $\int (3 x^5 - x^3 +6)\,dx= 3\int x^5\,dx -\int x^3\,dx +6\int 1\,dx$ Now we can find each indefinite integral using the reverse power rule: $\int x^n\,dx=\dfrac{x^{n+1}}{n+1}+C$ Note: we can only use the reverse power rule because $n \neq -1$. $\begin{aligned} &\phantom{=}\int (3 x^5 - x^3 +6)\,dx \\\\ &= 3\int x^5\,dx -\int x^3\,dx +6\int 1\,dx \\\\ &=3 \dfrac{x^6}{6} -\dfrac{x^4}{4} +6\dfrac{x^1}{1}+C \\\\ &=\dfrac{1}{2} x^6 -\dfrac{1}{4} x^4 +6 x+C \end{aligned}$ In conclusion, $\int (3 x^5 - x^3 +6)\,dx=\dfrac{1}{2} x^6 -\dfrac{1}{4} x^4 +6 x+C$